# Does an amphichiral knot admit roughly twice as many knot diagrams of a given crossing than a chiral knot?

Recall that a knot is amphichiral (or achiral) if there is a continuous deformation of the knot into its mirror image.

I'm interested in knowing when and whether we can use approaches like Stockmeyer counting to estimate and distinguish different knots. For example suppose that for most knots of grid dimension $$d$$, the total number of grid diagrams isotopic to a given knot is $$\mathcal O(n)$$. Then given two knot diagrams/grid diagrams $$K_1$$ and $$K_2$$ we can perform random Reidemeister moves on either $$K_1$$ or $$K_2$$ (allowing for a small increase in the grid dimension if needed), and hash the grid diagrams to do the counting. If $$K_1$$ and $$K_2$$ represent two distinct knots then there might be double $$\mathcal O(2n)$$ the number of such grid diagrams than if $$K_1$$ and $$K_2$$ represent the same knot.

Nonetheless if there's a sequence of such moves that transforms $$K_1$$ into a mirror image $$K_2$$ then the provided knot is amphichiral. But do we have any intuition on whether this would then mean that Stockmeyer counting would still say that the total number of grid diagrams of $$K_1$$ and $$K_2$$ together are twice as much as the total number of grid diagrams of another random knot $$K_0$$ of the same grid number $$d$$?

The Wikipedia article gives representative OEIS ID's for the number of amphichiral knots, and amphichirality appears rare.

I suspect this is not the case (although I don't have a proof of a formal statement). My heuristic is as follows. Start with a diagram $$D$$ for $$K$$ which is the same as its mirror. For negative amphichiral hyperbolic knots you can always find such a diagram (here is one for the figure eight knot - the symmetry is rotation around the marked point, then reflection across the plane of the diagram).
Now suppose you perform some Reidemeister moves to obtain a new diagram $$D'$$. Then indeed there is another diagram $$D''$$ for $$K$$ which is the mirror of $$D'$$, but since it can also be obtained from $$D$$ by a symmetric sequence of Reidemeister moves, we will already have counted it.
• I kind of envision the graph of Reidemeister moves as maybe having two large "lobes" - representing the two chiral halves. A walk on the graph - randomly tying and untying the knot - may or may not easily move between the two halves. In your heuristic can we be confident that the $D$ so provided - the one for which $K$ is the same as its mirror - is not in the middle of some bottleneck between the two chiral halves? For example, must a walk from one chirality to the other pass through the $D$ for which $K$ is the same as its mirror? May 21 at 23:43